Impulse Withstand Voltage Test

Because it wouldn't be an impulse if AC were applied!

Impulse voltage is a pulse voltage having time duration in micro seconds where as power frequency voltage is in millisecond.

Impulse voltage has a waveform approximately a Triangle having Vertex above base. The pulse peak voltage(Vp in Kv) rising time is T1(micro seconds) and fall or decaying time T2(micro seconds) is the time that falls to 50% of the peak rosed voltage.

Therefore a Impulse voltage is defined as a waveform of approximate Peak Voltage(Kv) with of the pulse duration by T2 and rising & falling time by ratio T1/T2.

This is what measured in Lightning stroke in thunder storms rainy season.

Since it a single pulse only. Hence we can not say exactly DC voltage. Because DC voltage has a constant average value for very long or infinite time.

The DC voltage is only the test level. The impulse should look like a square wave that only lasts for one cycle. The rate of change of the voltage from zero to the DC voltage level is the determining factor. Several frequencies are represented by that rate of change and consequently the ability of power to overcome the insulation is tested across a wide frequency range.

You can determine the frequency range by looking at the Laplace Transform of an impulse.

Dear Mr. NotUrordinary Joe

I can not say determination of the frequency range by looking at the Laplace Transform of an impulse.

I can definitely say to use curve fitting to represent impulse wave as

V = a+b*t +c*t² + d*t³+e*t⁴+f*t⁵+g*t⁶+h*t⁷+j*t⁸+k*t⁹

^{Where: V= Voltage in KV and t= Time in microseconds}

^{a, b, c ,d, e, f, g, h, j, and k are constants which can be determined V equation. So you have 10 equations and 10 constants which can be easily calculated by picking 10 important point on the curve such as initial, peak, end on the curve. Equation of Voltage V can easily represented by in terms of Time with all 10 constants.}

^{Impulse wave form is shown as below:}